Question

我有一个User类。还有两个子类。父母和孩子。我使用{“user”：“...”}从我的服务器获取json，需要将其转换为父级或子级，具体取决于user.type

据我所知，我需要以这种方式添加自定义转换器：

        Moshi moshi = new Moshi.Builder()
            .add(new UserAdapter())
            .build();

这是我对UserAdapter的实现。我知道这是假的，但它甚至不能这样工作：

public class UserAdapter {

@FromJson
User fromJson(String userJson) {
    Moshi moshi = new Moshi.Builder().build();
    try {
        JSONObject jsonObject = new JSONObject(userJson);
        String accountType = jsonObject.getString("type");

        switch (accountType) {
            case "Child":
                JsonAdapter<Child> childJsonAdapter = moshi.adapter(Child.class);
                return childJsonAdapter.fromJson(userJson);
            case "Parent":
                JsonAdapter<Parent> parentJsonAdapter = moshi.adapter(Parent.class);
                return parentJsonAdapter.fromJson(userJson);

        }
    } catch (JSONException | IOException e) {
        e.printStackTrace();
    }

    return null;
}

@ToJson
String toJson(User user) {
    Moshi moshi = new Moshi.Builder().build();
    JsonAdapter<User> jsonAdapter = moshi.adapter(User.class);
    String toJson = jsonAdapter.toJson(user);
    return toJson;
}

首先，我使用此代码获得以下异常。

com.squareup.moshi.JsonDataException: Expected a string but was BEGIN_OBJECT at path $.user

其次，我相信有更好的方法。请指教。

UPD。这是错误的堆栈跟踪：

 com.squareup.moshi.JsonDataException: Expected a name but was BEGIN_OBJECT at path $.user
 at com.squareup.moshi.JsonReader.nextName(JsonReader.java:782)
 at com.squareup.moshi.ClassJsonAdapter.fromJson(ClassJsonAdapter.java:141)
 at com.squareup.moshi.JsonAdapter$1.fromJson(JsonAdapter.java:68)
 at com.squareup.moshi.JsonAdapter.fromJson(JsonAdapter.java:33)
 at retrofit.MoshiResponseBodyConverter.convert(MoshiResponseBodyConverter.java:33)
 at retrofit.MoshiResponseBodyConverter.convert(MoshiResponseBodyConverter.java:23)
 at retrofit.OkHttpCall.parseResponse(OkHttpCall.java:148)
 at retrofit.OkHttpCall.execute(OkHttpCall.java:116)
 at retrofit.RxJavaCallAdapterFactory$CallOnSubscribe.call(RxJavaCallAdapterFactory.java:111)
 at retrofit.RxJavaCallAdapterFactory$CallOnSubscribe.call(RxJavaCallAdapterFactory.java:88)
 at rx.Observable$2.call(Observable.java:162)
 at rx.Observable$2.call(Observable.java:154)
 at rx.Observable$2.call(Observable.java:162)
 at rx.Observable$2.call(Observable.java:154)
 at rx.Observable.unsafeSubscribe(Observable.java:7710)
 at rx.internal.operators.OperatorSubscribeOn$1$1.call(OperatorSubscribeOn.java:62)
 at rx.internal.schedulers.ScheduledAction.run(ScheduledAction.java:55)
 at java.util.concurrent.Executors$RunnableAdapter.call(Executors.java:422)
 at java.util.concurrent.FutureTask.run(FutureTask.java:237)
 at java.util.concurrent.ScheduledThreadPoolExecutor$ScheduledFutureTask.access$201(ScheduledThreadPoolExecutor.java:152)
 at java.util.concurrent.ScheduledThreadPoolExecutor$ScheduledFutureTask.run(ScheduledThreadPoolExecutor.java:265)
 at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1112)
 at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:587)
 at java.lang.Thread.run(Thread.java:818)

Answer 1

在我看来，您希望自定义de /序列化JSON数据的示例如下：https://github.com/square/moshi#another-example

它使用与JSON结构相对应的中间类，并且Moshi会自动为您充气。然后，您可以使用膨胀的数据来构建专门的用户类。例如：

// Intermediate class with JSON structure
class UserJson {
  // Common JSON fields
  public String type;
  public String name;
  // Parent JSON fields
  public String occupation;
  public Long salary;
  // Child JSON fields
  public String favorite_toy;
  public Integer grade;
}

abstract class User {
  public String type;
  public String name;
}

final class Parent extends User {
  public String occupation;
  public Long salary;
}

final class Child extends User {
  public String favoriteToy;
  public Integer grade;
}

现在，适配器：

class UserAdapter {
  // Note that you pass in a `UserJson` object here
  @FromJson User fromJson(UserJson userJson) {
    switch (userJson.type) {
    case "Parent":
      final Parent parent = new Parent();
      parent.type = userJson.type;
      parent.name = userJson.name;
      parent.occupation = userJson.occupation;
      parent.salary = userJson.salary;
      return parent;
    case "Child":
      final Child child = new Child();
      child.type = userJson.type;
      child.name = userJson.name;
      child.favoriteToy = userJson.favorite_toy;
      child.grade = userJson.grade;
      return child;
    default:
      return null;
    }
  }

  // Note that you return a `UserJson` object here.
  @ToJson UserJson toJson(User user) {
    final UserJson json = new UserJson();
    if (user instanceof Parent) {
      json.type = "Parent";
      json.occupation = ((Parent) user).occupation;
      json.salary = ((Parent) user).salary;
    } else {
      json.type = "Child";
      json.favorite_toy = ((Child) user).favoriteToy;
      json.grade = ((Child) user).grade;
    }
    json.name = user.name;
    return json;
  }
}

我认为这更清晰，并允许Moshi做它的事情，即从JSON创建对象并从对象创建JSON。没有老式的JSONObject！

测试：

Child child = new Child();
child.type = "Child";
child.name = "Foo";
child.favoriteToy = "java";
child.grade = 2;
Moshi moshi = new Moshi.Builder().add(new UserAdapter()).build();
try {
  // Serialize
  JsonAdapter<User> adapter = moshi.adapter(User.class);
  String json = adapter.toJson(child);
  System.out.println(json);
  // Output is: {"favorite_toy":"java","grade":2,"name":"Foo","type":"Child"}

  // Deserialize
  // Note the cast to `Child`, since this adapter returns `User` otherwise.
  Child child2 = (Child) adapter.fromJson(json);
  System.out.println(child2.name);
  // Output is: Foo
} catch (IOException e) {
  e.printStackTrace();
}

Answer 2

您可能尝试根据以下内容实现解析：https://github.com/square/moshi#custom-type-adapters

String被用作@FromJson方法的参数，因此它可以被神奇地解析为一些映射助手类或String，我们必须手动解析它，对吧？实际上没有，您可以使用映射辅助类或映射。

因此，您的异常Expected a string but was BEGIN_OBJECT at path $.user是由Moshi尝试将该用户作为String（因为这是您在适配器中隐含的内容）引起的，而它只是另一个对象。

我不喜欢将所有可能的字段解析为某个帮助器类，因为在多态性的情况下，类可能变得非常大并且您需要依赖或记住/注释代码。

您可以将其作为地图处理 - 这是未知类型的默认模型 - 并将其转换为json，因此在您的情况下看起来像：

    @FromJson
    User fromJson(Map<String, String> map) {
        Moshi moshi = new Moshi.Builder().build();
        String userJson = moshi.adapter(Map.class).toJson(map);
        try {
            JSONObject jsonObject = new JSONObject(userJson);
            String accountType = jsonObject.getString("type");

            switch (accountType) {
                case "Child":
                    JsonAdapter<Child> childJsonAdapter = moshi.adapter(Child.class);
                    return childJsonAdapter.fromJson(userJson);
                case "Parent":
                    JsonAdapter<Parent> parentJsonAdapter = moshi.adapter(Parent.class);
                    return parentJsonAdapter.fromJson(userJson);

            }
        } catch (JSONException | IOException e) {
            e.printStackTrace();
        }

        return null;
    }

当然你可以直接处理map：检索“type”字符串，然后将剩下的map解析为所选的类。然后根本没有必要使用JSONObject，并且不依赖于Android并且更容易测试解析。

    @FromJson
    User fromJson(Map<String, String> map) {
        Moshi moshi = new Moshi.Builder().build();
        try {
            String userJson = moshi.adapter(Map.class).toJson(map);
            switch (map.get("type")) {
                case "Child":
                    JsonAdapter<Child> childJsonAdapter = moshi.adapter(Child.class);
                    return childJsonAdapter.fromJson(userJson);
                case "Parent":
                    JsonAdapter<Parent> parentJsonAdapter = moshi.adapter(Parent.class);
                    return parentJsonAdapter.fromJson(userJson);

            }
        } catch (IOException e) {
            e.printStackTrace();
        }

        return null;
    }

Answer 3

现在有了使用PolymorphicJsonAdapterFactory的更好的方法。参见https://proandroiddev.com/moshi-polymorphic-adapter-is-d25deebbd7c5

定制转换器与Moshi子类

3 个答案: